Prove that $A^k = 0 $ iff $A^2 = 0$ Let $A$ be a $  2 \times  2 $ matrix and  a positive integer $k \geq 2$. Prove that $A^k = 0 $ iff $A^2 = 0$.
I can make it to do this exercise if I have $ \det (A^k) = (\det A)^k $. But this question comes before this.
Thank you very much for your help!
 A: One direction is trivial, so let's consider the other.
I will give you some hints towards a solution.
Suppose $A^k = 0$. This means that $A$ satisfies the polynomial $x^k$. What does this say about the minimal polynomial of the matrix? What in turn does this say about the characteristic polynomial? Finally, consider the Cayley-Hamilton theorem.
A: The operator $A$ is nilpotent, therefore there exists a basis where $A$ is strictly upper triangular by Engel's Theorem. It follows that $A^2 = 0$.
A: The solution using the minimal polynomial and Cayley-Hamilton is a bit of an over-kill and somewhat of a magic solution. I prefer the following non-magic solution (for the non-trivial implication, and there is no need to assume the matrix is $2\times 2$). 
Think of $A$ as a linear operator $A: V \to V$ with $V$ an $n$-dimensional vector space, and suppose $A^t=0$. We'll show that $A^n=0$. Now, for each $m\ge 1$ consider the space $K_m$, the kernel of $A^m$. It is immediate that $K_{m}\subseteq K_{m+1}$. Since $A^t=0$ it follows that $K_t=V$. It is also immediate that if $K_m=K_{m+1}$, then $K_m=K_{r}$ for all $r>m$. 
Thus, the sequence of kernels is an increasing sequence of subspaces, that stabilizes as soon as the next step is equal to the previous one, and it is eventually all of $V$. Now we will use the fact that $V$ is of dimension $n$. Considering the dimensions of the kernels, the above implies that the sequence of dimensions is strictly increasing until it stabilizes. Since it reaches $V$, the dimensions reach $n$ and since the dimension of $K_1$ is not zero, the sequence of dimensions starts at $1$ or more. That means it has to stabilize after no more than $n$ steps, and thus $K_t=V$ for some $t\le n$. But then $K_n=V$, and thus $A^n=0$.
A: The standard and more general approach to solve this kind of problems is to consider the minimal polynomial of $A$ (see the other answers here). However, in this particular problem, as $A$ is only $2\times2$, one may solve the problem as follows. As the forward implication is trivial, we consider only the backward implication
Proof 1. If $A^k=0$ for some $k\ge2$, then $A$ is singular and it has at most rank 1. Therefore, $A=uv^T$ for some vectors $u$ and $v$. Note that for any $n\ge1$,
$$
A^n=\underbrace{(uv^T)(uv^T)\cdots(uv^T)}_{n \text{ times}}
=u\underbrace{(v^Tu)}_{\text{scalar}}\cdots(v^Tu)v^T=(v^Tu)^{n-1}A.
$$
In particular, $A^k=(v^Tu)^{k-1}A$ and $A^2=(v^Tu)A$. Now the result follows immediately.
Proof 2. If $A^k=0$ for some $k\ge2$, then $A$ is singular.
Let $Ax=0$ for some $x\ne0$. Extend $x$ to a basis $\{x,y\}$ of the underlying vector space. Let $Ay=px+qy$. By mathematical induction, we have $A^ny=pq^{n-1}x + q^ny$ for any $n\ge1$. Now use $A^k=0$ to argue that $q=0$. Hence $A^2x=A^2y=0$.
A: Here's a naive solution that uses only the definition of linear independence, the notion of dimension, and the fact that $\dim \Bbb F^2 = 2$.
One direction is trivial. For the other direction, for clarity of presentation we'll give an explicit argument for $k = 3$, but the general case is totally analogous:
Suppose $A^2 \neq 0$; then, there is a vector $\xi \in \Bbb F^2$ such that $A^2 \xi \neq 0$. We'll show that the triple $(\xi, A \xi, A^2 \xi)$ is linearly independent, which implies that $\dim \Bbb F^2 \geq 3$, a contradiction:
Suppose not; then, there are coefficients $a, b, c$, not all $0$, such that $$0 = a \xi + b A \xi + c A^2 \xi.$$ Multiplying both sides of this equation by $A^2$ gives
$$0 = a A^2 \xi + b A^3 \xi + c A^4 \xi = a A^2 \xi$$ (the second equality uses that $A^3 = 0$). Since $A^2 \xi \neq 0$, we must have $a = 0$ and so $$0 = b A \xi + c A^2 \xi .$$ Multiplying both sides by $A$ similarly leads to $b = 0$, leaving $$c A^2 \xi = 0,$$ so $c = 0$, a contradiction.
A: One direction is clear.
Suppose $A^k = 0$ . This means that the minimal polynomial of $A$, call it $g(x)$, satisfies $$g(x) \mid x^k$$ Thus $$g(x) = x^n \ \ \ n \leq k$$ But the minimal polynomial divides the characteristic polynomial, which has degree $2$, and so $n = 1$ or $n = 2$. In both cases $$A^2 = 0$$
A: A^k=0 iff A^2=0  k positive interger >2 you can write k=2+n and n>=1. Using induction: A^2+n=0
Then A^2=0. Take A^3=A^2A=0. And  prove  A^2+n+1=0 =>A^2=0  and A^2+nA=0.  Then A^2=0.
A: A^k=0 iff A^2=0 k positive interger >2 you can write k=2+n and n>=1. Using induction: A^2+n=0 Then A^2=0. Take A^3=A^2A=0. And prove A^2+n+1=0 =>A^2=0 and A^2+nA=0. Then A^2=0.
First direction A^2=0 then A^k=0. 
You can take A and B as a matrix.. Lets represent both matrix . We have A=[a b over c d] and B=[b1 b2 over b3 b4]  then we have B times A=0.
 (multiplying) we have
BA= (b1a+b2c  and  b1b+ b2d  over b3a+b4c and b3b+b4d)
It was the matrix. Then
b1a+b2c=0   b1b+ b2d=0  b3a+b4c=0  b3b+b4d=0
Where we have
(ad-bc)b1=0   (ad-bc)b2=0  (ad-bc)b3=0 and
 (ad-bc)b4=0
Then we have 2 possibilities:
b1=b2=b3=b4=0 or ad-bc=0 this is B=0 or ad=bc
Obviously if A^2=0 then A^k =0 for every k bigger than 2.
A second direction A^k=0 then A^2=0 :
Now let's prove A^k=0  for k>2 with k as positive integer. Then A^2=0.
Watch first k>2, then you can write k=2+n with k positive integer  n>=1.
Let's prove using induction that: 
 if A^2+n=0 then A^2=0
For n=1 we have A^3=A^2.A=0. 
Previously we worked on matrix we can say A^2=0 or ad=bc. 
Let's represent A^3 as matrix:
A^3=[(a^2+bc)a+(ab+bd)c   and  (a^2+bc)b+(ab+bd)d  over (ca+dc)a+(bc+d^2)c  and   (ca+dc)b+(bc+d^2)d)]
It was the matrix. 
A^3=0 then 
(a^2+bc)(a+d)=0    (ab+bc)(a+d)=0
 (ca+dc)(a+d)=0    (bc+d^2)(a+d)=0
Then  let's represent the matrix as : A^2=[a^2+bc  and  ab+bd  over
ca+dc   and  bc+d^2]=[0  and 0  over 0. And 0]
Then 
a=-d but we have ad=bc  then a^2+bc=0 
bc+d^2=0  ab+bd=0  and ca+dc=0 then A^2=0
And prove that A^2+n+1=0 then A^2=0
If A^2+n.A=0 then A^2+n.A=0. Because previous work  A^2+n=0 or ad=bc
 (induction) we conclude A^2+n=0 and 
Because the induction hypothesis we have A^2=0 finally we have A^k=0 then A^2=0. Done
